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aceace Group TitleBest ResponseYou've already chosen the best response.0
Find a relation between x and y that does NOT involve logarithms: log x + log y = log (x+ y)
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
Can we assume it's a log of base 10?
 2 years ago

aceace Group TitleBest ResponseYou've already chosen the best response.0
yes i think so
 2 years ago

AERONIK Group TitleBest ResponseYou've already chosen the best response.0
logx + log y is never log(x+y) it seems... it is log(xy)
 2 years ago

aceace Group TitleBest ResponseYou've already chosen the best response.0
do you want the answer?
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
I suppose we could exponentiate both sides.. \[\huge 10^{(\log x + \log y)}=10^{\log(x+y)}\] Using rules of exponents we can split up the left side like this: \[\huge 10^{\log x}*10^{\log y}=10^{\log(x+y)}\] Recognizing that log base 10, and the exponentiation base 10 are inverse operations of one another, they essentially "cancel out". Giving us: \[\huge xy=(x+y)\] \[\huge y=\frac{x}{x1}\]
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
Something like that maybe? :o
 2 years ago

geoffb Group TitleBest ResponseYou've already chosen the best response.0
@zepdrix Oooh, sexy!
 2 years ago

aceace Group TitleBest ResponseYou've already chosen the best response.0
@zeptr yes thats the answer !!!
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
Oh cool. :)
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
Were you just testing us or something? :D lol Knew the answer the whole time huh? XD
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
@AERONIK , it seems that logx+logy=log(x+y) sometimes. I wouldn't say NEVER. Obviously it doesn't match the rule for logs that we had in mind. But if you plug in y=2, x=2 into the original equation, that should work out ok as one solution :D Since 2+2 is the same as 2*2
 2 years ago

aceace Group TitleBest ResponseYou've already chosen the best response.0
no i had the answers at the back of book but i have no idea how to do it... i still dont get the logic though...
 2 years ago

AERONIK Group TitleBest ResponseYou've already chosen the best response.0
so it is not an identity and just a function you say ?
 2 years ago

aceace Group TitleBest ResponseYou've already chosen the best response.0
yeh just a random equation
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
Yah it appears it wasn't testing you on whether or not you knew your log identities, they just wanted to see a relationship between x and y. Interesting problem :O
 2 years ago

AERONIK Group TitleBest ResponseYou've already chosen the best response.0
and one more thing in your proof you have assumed the given statement to be correct, can you please give a formal proof for te same?
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
what now? :o
 2 years ago

AERONIK Group TitleBest ResponseYou've already chosen the best response.0
and is it log to the base 10 or log to the base e ?
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
For this problem, we have to assume that they are ALL the same base. Otherwise it won't work out the same. If they're a different base, let's say A, then we exponentiate, writing each side with a base A (regardless of what A might be) and it will work out the same! :D
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
Different than 10 i mean*
 2 years ago

AERONIK Group TitleBest ResponseYou've already chosen the best response.0
oh ok. sorry ..
 2 years ago

aceace Group TitleBest ResponseYou've already chosen the best response.0
10 logx ∗10 logy =10 log(x+y) Recognizing that log base 10, and the exponentiation base 10 are inverse operations of one another, they essentially "cancel out". HOW DID YOU DO THIS STEP... CAN YOU SHOW ME? PLEASE
 2 years ago

AERONIK Group TitleBest ResponseYou've already chosen the best response.0
by defination if y=log x to the base a, then x is nothing but a^y, just use this and you can prove the identity.
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
\[\large \arcsin( \sin x)=x\]\[\huge e^{\ln x}=x\] Yah this is true of any function and it's inverse :D Remember it in trig with the inverse functions?
 2 years ago

aceace Group TitleBest ResponseYou've already chosen the best response.0
no ...can you do a proof for it please?
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.3
dw:1353225574033:dw Hmmm this might be a little confusing.. I drew a lot of arrows : lemme know if it makes some sense.
 2 years ago

aceace Group TitleBest ResponseYou've already chosen the best response.0
i get it ...thanks so much
 2 years ago
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